我要成為百度之星
阿新 • • 發佈:2019-01-12
http://oj.acm.zstu.edu.cn/JudgeOnline/problem.php?id=4233
題解:題目和解答完全不一樣,答案是101010101010101010......的字串
題意是和二進位制位有關的字串排列。
某一位的狀態和這一位的序號-1的二進位制中1個數的奇偶有關
C++版本一
AC版本
/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp; ll l,r; int main() { #ifdef DEBUG freopen("input.in", "r", stdin); //freopen("output.out", "w", stdout); #endif scanf("%d",&t); while(t--){ scanf("%lld%lld",&l,&r); if(l>r){ swap(l,r); } temp=0; if(l%2==0){ l++; } if(r%2==1){ temp++; r--; } printf("%lld\n",(r-l+1)/2+temp); } //cout << "Hello world!" << endl; return 0; }
C++版本二
AC版本
#include<stdio.h> #include<algorithm> #include<string.h> #include<math.h> #include<vector> using namespace std; vector<int> V; int solve(int x) { if(x==0) return 0; if(x==1) return 1; int y=lower_bound(V.begin(),V.end(),x)-V.begin(); if(V[y]==x) return x/2; int dif=V[y]-x; return solve(dif)+(x-dif)/2; } int main() { V.push_back(2); int x=2; while(x<=(int)1e9) { x=x*2; V.push_back(x); } int T; scanf("%d",&T); while(T--) { int l,r; scanf("%d%d",&l,&r); printf("%d\n",solve(r)-solve(l-1)); } return 0; }
C++版本三
符合題意版本
/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp; ll l,r; int main() { #ifdef DEBUG freopen("input.in", "r", stdin); //freopen("output.out", "w", stdout); #endif //scanf("%d",&n); scanf("%d",&t); while(t--){ scanf("%lld%lld",&l,&r); if(l>r){ swap(l,r); } if(l%2==0){ if(__builtin_popcount(l-1)%2==0) l--; else l++; } if(r%2==1){ if(__builtin_popcount(r-1)%2==0) r++; else r--; } printf("%lld\n",(r-l+1)/2); } //cout << "Hello world!" << endl; return 0; }